It seems that coding tests are conducted overseas in interviews with engineers, and in many cases, the main thing is to implement specific functions and classes according to the theme.
Apparently, many engineers take measures on the site called LetCode.
It is a site that trains the algorithmic power that can withstand the coding test that is being done in the early story, and it is an inevitable path for those who want to build a career at an overseas tech company.
I wrote it in a big way, but I have no plans to have such an interview at the moment.
However, as an IT engineer, it would be better to have the same level of algorithm power as a person, so I would like to solve the problem irregularly and write down the method I thought at that time as a memo.
I'm solving it with Python3.
Leet Code Table of Contents Starting from Zero
Last time Leet Code Day74 starting from zero "12. Integer to Roman"
Right now, I'm prioritizing the Medium of the Top 100 Liked Questions. I solved all Easy, so if you are interested, please go to the table of contents.
Twitter I'm doing it.
** Technical Blog Started! !! ** ** I think the technology will write about LetCode, Django, Nuxt, and so on. ** This is faster to update **, so please bookmark it!
15. 3Sum The difficulty level is Medium.
The problem is, if you have an array of n integers nums, then you have an algorithm to check if nums has elements ʻa, b, c such that ʻa + b + c = 0. Design and find all the unique triplets in an array that gives the sum of 0s.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
You can choose 3 and choose the combination that will be 0. It may be said that it is an iron rule of combination, but I think it is better to decide what to fix first.
This time I wrote as follows.
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
ans = []
for i in range(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
left = i+1
right = len(nums)-1
while left < right:
sums = nums[i]+nums[left]+nums[right]
if sums < 0:
left += 1
elif sums > 0:
right -= 1
else:
ans.append([nums[i],nums[left],nums[right]])
while left < right and nums[left]==nums[left+1]:
left += 1
while left > right and nums[right]==nums[right-1]:
right -= 1
left += 1
right -= 1
return ans
# Runtime: 1092 ms, faster than 45.79% of Python3 online submissions for 3Sum.
# Memory Usage: 17.1 MB, less than 76.15% of Python3 online submissions for 3Sum.
Sort in ascending order by sorting first. This makes it easier to specify the size of the element when turning the element with a for statement from before.
To give a concrete example,
[-1,0,1,2,-1,-4]
When sorted,
[-4,-1,-1,0,1,2]
It will be like that.
The core of this problem is how to select the remaining two elements of 3sum when licking the elements of this from the beginning.
As I wrote in the code, I set ʻi + 1toleft, len (nums) -1toright, and if they were added, their total value, sums`, would be 0. If it is smaller, the left element is +1. If it is larger than 0, right is -1.
This is possible because of the sorting mentioned above.
actually
-4+(-1)+2=-3<0
So move the left element and
-4+(-1)+2=-3<0
This is still the same, so move the left element again
-4+0+2=-2<0
It is a flow like that.
And if other than those, that is, sums == 0, the element is added to the prepared ʻans`.
I feel that it is easy to solve such a combination system of sums in a format like this one, so I would like to remember it as a typical problem.
So that's it for this time. Thank you for your hard work.
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